Ah, yes. It is Water and Wastewater Engineering: Design Principles and Practice Mackenzie L Davis Published by McGraw-Hill Education, 2013 ISBN 10: 1259064832 / ISBN 13: 9781259064838
I didn't use very many of the slides in this video, but here is the file with all the slides from my coagulation/flocculation lecture: cecas.clemson.edu/ladnergroup/share/Chapter3-Coagulation-Flocculation.pptx
Dear Sir , Mr David Ladner , Really thank you for your very comprehensive explanation and very much helpful for us . How can we calculate the Gradient G value ?
this video is super helpful. I was wondering if these same relations are applicable to parallel flow flocculators. Are there different guidelines for this latter ser up? (i.e. relative speed coeff - drag coeff - tapered flocculation - cross sectional %age area) Thank you!!!!
Sorry for the very slowly reply; I overlooked this. In general, the concepts should be applicable, yes. And I would think the G values and GT values could be the same in either design. But I haven't delved into parallel flow designs, so there could be some caveats. Plus, in real treatment plants there is a lot of variability, so if I were designing something I would go to similar facilities and see what they are doing.
how much percentage cross sectional area is considered good? could you please specify the range? like we got 17% here and in my design I got 14%. Please specify the range in which it is considered favorable. Thank you Sir.
I didn't have it posted anywhere previously, but I just uploaded it to cecas.clemson.edu/ladnergroup/share/Flocculation-Example.pdf Hope it is helpful.
I was struggling to calculate RPM. This really helped. Thanks!
How come in tip speed, 3.15 X 0.0875=0.87m/s? It should be 0.275m/s..
very nice information.Thanks.
this really help. thank you
How come in tip speed, 3.15 X 0.0875=0.87m/s? It should be 0.275m/s..
Thank you so much for this helpful video. Could you also share with us what textbook you are using for this class?
Ah, yes. It is Water and Wastewater Engineering: Design Principles and Practice
Mackenzie L Davis
Published by McGraw-Hill Education, 2013
ISBN 10: 1259064832 / ISBN 13: 9781259064838
@@davidladner Thank you for sharing this information. God bless!
This was a really comprehensive example thank you. Are the slides used in this presentation available anywhere?
I didn't use very many of the slides in this video, but here is the file with all the slides from my coagulation/flocculation lecture: cecas.clemson.edu/ladnergroup/share/Chapter3-Coagulation-Flocculation.pptx
David Ladner thank you so much.
Dear Sir , Do you have the On line Training course ? We would like to attend .
Dear Sir , Mr David Ladner , Really thank you for your very comprehensive explanation and very much helpful for us . How can we calculate the Gradient G value ?
this video is super helpful. I was wondering if these same relations are applicable to parallel flow flocculators.
Are there different guidelines for this latter ser up? (i.e. relative speed coeff - drag coeff - tapered flocculation - cross sectional %age area)
Thank you!!!!
Sorry for the very slowly reply; I overlooked this. In general, the concepts should be applicable, yes. And I would think the G values and GT values could be the same in either design. But I haven't delved into parallel flow designs, so there could be some caveats. Plus, in real treatment plants there is a lot of variability, so if I were designing something I would go to similar facilities and see what they are doing.
how much percentage cross sectional area is considered good? could you please specify the range? like we got 17% here and in my design I got 14%. Please specify the range in which it is considered favorable. Thank you Sir.
please where i can catch this example document PDF or ntp
I didn't have it posted anywhere previously, but I just uploaded it to cecas.clemson.edu/ladnergroup/share/Flocculation-Example.pdf
Hope it is helpful.
thank you so much
Thank you very much, the example is very useful. have some example to design a sedimentation tank
How come in tip speed, 3.15 X 0.0875=0.87m/s? It should be 0.275m/s..
@@saurajitsinha2852 it is pi*3.15*0.0875